Advanced function
Question 13
By; Nada and Ayah
2021-10-25
(B)Q13) A polynomial has a single root at x=1, a double root at x=-2 and two complex Roots at
x= -2 + i and x= 1+i . If this polynomial has a y-intercept at (0,80) determine the equation of this polynomial
Our Question
procedure
The (x)
Calculations
x=1
x=-2
x=-2+i
x= -2-i (Conjugate)
x= 1+i
x=1-i (Conjugate)
y=a(x+2-i)(x+2+i)(x-1-i)(x-1+i)(x-1)(x+2)^2
y=a([(x+2)^2 -i^2][(x-1)^2-i^2](x-1)(x+2)^2
y=a([(x+2)^2 +1][(x-1)^2+1](x-1)(x+2)^2
y=a([(x^2+4x+4)+1][(x^2-2x+1)+1](x-1)(x+2)^2
y=a((x^2+4x+5)(x^2-2x+2)(x-1)(x+2)^2
Final Eqaution
Finding (a)
For the function
For the function
Use (0,80) which is the y-intercept
y=a(x-1)(x+2)^2(x^2-2x+2)(x^2+4x+5) We plugged in our y intercept as 80 and our (x) as zero 80=a(0-1)(0+2)^2(0^2-2(0)+2) (0^2+4(0)+5) 80=a(-1)(4)(2)(5)
80=a(-40)
80 =a -40 a= -2
By plugging our a=-2 we recive a final equation as y=-2(x-1)(x+2)^2(x^2-2x+2)(x^2+4x+5)
Thanks for you listening
Any question?
Advanced Function
ayah benzaibek
Created on October 25, 2021
Start designing with a free template
Discover more than 1500 professional designs like these:
View
Audio tutorial
View
Pechakucha Presentation
View
Desktop Workspace
View
Decades Presentation
View
Psychology Presentation
View
Medical Dna Presentation
View
Geometric Project Presentation
Explore all templates
Transcript
Advanced function
Question 13
By; Nada and Ayah
2021-10-25
(B)Q13) A polynomial has a single root at x=1, a double root at x=-2 and two complex Roots at x= -2 + i and x= 1+i . If this polynomial has a y-intercept at (0,80) determine the equation of this polynomial
Our Question
procedure
The (x)
Calculations
x=1 x=-2 x=-2+i x= -2-i (Conjugate) x= 1+i x=1-i (Conjugate)
y=a(x+2-i)(x+2+i)(x-1-i)(x-1+i)(x-1)(x+2)^2 y=a([(x+2)^2 -i^2][(x-1)^2-i^2](x-1)(x+2)^2 y=a([(x+2)^2 +1][(x-1)^2+1](x-1)(x+2)^2 y=a([(x^2+4x+4)+1][(x^2-2x+1)+1](x-1)(x+2)^2 y=a((x^2+4x+5)(x^2-2x+2)(x-1)(x+2)^2
Final Eqaution
Finding (a)
For the function
For the function
Use (0,80) which is the y-intercept y=a(x-1)(x+2)^2(x^2-2x+2)(x^2+4x+5) We plugged in our y intercept as 80 and our (x) as zero 80=a(0-1)(0+2)^2(0^2-2(0)+2) (0^2+4(0)+5) 80=a(-1)(4)(2)(5) 80=a(-40) 80 =a -40 a= -2
By plugging our a=-2 we recive a final equation as y=-2(x-1)(x+2)^2(x^2-2x+2)(x^2+4x+5)
Thanks for you listening
Any question?